Light with a wavelength in vacuum of 514 nm falls perpendicularly on a biologica

Question

Light with a wavelength in vacuum of 514 nm falls perpendicularly on a biological specimen that is 2.00 m thick. The light splits into two beams polarized at right angles, for which the indices of refraction are n1 = 1.320 and n2 = 1.333, respectively.

(a) Calculate the wavelength of each component of the light while it is traversing the specimen. (Give your answers to at least one decimal place.)

(b) Calculate the phase difference between the two beams when they emerge from the specimen.
______ °

Explanation / Answer

a) lambda new = lambda old* refractive index
lambda n1 = 514/1.320 = 389.39 nm
lambda n2 = 514/1.333 = 385.59 nm

b) Number of cycles completed by wavelength 1 in the specimen = 2*10^-6/389.39*10^-9 = 5.136
Number of cycles completed by wavelength 2 in the specimen = 2*10^-6/385.59*10^-9 = 5.186
Cycle difference = 0.05
1 cycle difference = 2*pi phanse diff
Phase diff = 2*pi*0.05 = 0.319 rad

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